# proof of properties of derivatives by pure algebra

###### Theorem 1.

The derivative satisfies the following rules:

• Linearity
 $\frac{d}{dx}(f(x)+g(x))=\frac{df}{dx}+\frac{dg}{dx},\qquad\frac{d}{dx}(af(x))=% a\frac{df}{dx},$

for $f(x),g(x)\in R[x]$ and $a\in R$.

•  $\frac{d}{dx}(x^{n})=nx^{n-1}.$
•  $\frac{d}{dx}(f(x)g(x))=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}.$
###### Remark 2.

The following proofs apply to derivatives by pure algebra (http://planetmath.org/DerivativesByPureAlgebra). While the nature of the proofs are similar to the usual proofs, the notion of a limit is replaced by modular arithmetic in $R[x,h]/(h)$.

###### Proof.

Power rule.

 $\displaystyle\frac{d}{dx}(x^{n})$ $\displaystyle\equiv$ $\displaystyle\frac{(x+h)^{n}-x^{n}}{h}$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{n}\binom{i}{j}x^{n-j}h^{j-1}$ $\displaystyle\equiv$ $\displaystyle\binom{n}{1}x^{n-1}=nx^{n-1}.$

Linearity rule. For all $f(x),g(x)\in R[x]\cong R[x,h]/(h)$, it follows

 $\frac{(f+g)(x+h)-(f+g)(x)}{h}\equiv\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\equiv% \frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}.$

Furthermore, for all $a\in R$

 $\frac{(af)(x+h)-(af)(x)}{h}\equiv\frac{af(x+h)-af(x)}{h}=a\frac{f(x+h)-f(x)}{h}.$

Product rule. In $R[x,h]$ modulo $(h)$ we have:

 $\displaystyle\frac{d}{dx}(fg)$ $\displaystyle\equiv$ $\displaystyle\frac{f(x+h)g(x+h))-f(x)g(x)}{h}$ $\displaystyle\equiv$ $\displaystyle\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}$ $\displaystyle\equiv$ $\displaystyle\frac{(f(x+h)-f(x))g(x+h)+f(x)(g(x+h)-g(x))}{h}$ $\displaystyle\equiv$ $\displaystyle\frac{f(x+h)-f(x)}{h}g(x+h)+f(x)\frac{g(x+h)-g(x)}{h}$ $\displaystyle\equiv$ $\displaystyle\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}.$

Title proof of properties of derivatives by pure algebra ProofOfPropertiesOfDerivativesByPureAlgebra 2013-03-22 16:00:03 2013-03-22 16:00:03 Algeboy (12884) Algeboy (12884) 5 Algeboy (12884) Proof msc 26B05 msc 46G05 msc 26A24 RulesOfCalculusForDerivativeOfPolynomial